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cos^6x=cos^4x

Guest Nov 8, 2014

Best Answer 

 #2
avatar+87333 
+10

cos^6x = cos^4x     subtract cos^4x ffrom both sides

cos^6x - cos^4x =  0    faactor this

cos^4x*(cos^2x - 1) = 0

cos^4x (cosx + 1)(cosx-1)   setting each factor to 0, we have

x = 0,  pi/2,  pi , (3 pi)/2        in the interval  [0, 2pi)

More general solutions are  x = 0 + (pi)n  and   x = pi/2 + (pi/2)n    where n is an integer

 

CPhill  Nov 8, 2014
 #1
avatar+92806 
+5

$$\\cos^6x=cos^4x\\
$Divide both sides by $\;cos^4x\\
cos^2x=1\\
cosx=\pm1
$At this point I just think about the unit circle$\\
$cos of an angle is given by the x value (not the angle x)$\\
$The x value is 1 at 0 and it is -1 at 180 degrees$\\
So \\
x=180n$ degrees where $ n\in Z\\
or\\
x=n\pi$ radians where $ n\in Z\\$$

Melody  Nov 8, 2014
 #2
avatar+87333 
+10
Best Answer

cos^6x = cos^4x     subtract cos^4x ffrom both sides

cos^6x - cos^4x =  0    faactor this

cos^4x*(cos^2x - 1) = 0

cos^4x (cosx + 1)(cosx-1)   setting each factor to 0, we have

x = 0,  pi/2,  pi , (3 pi)/2        in the interval  [0, 2pi)

More general solutions are  x = 0 + (pi)n  and   x = pi/2 + (pi/2)n    where n is an integer

 

CPhill  Nov 8, 2014
 #3
avatar+92806 
+5

Okay I missed some answers - thanks for picking that up chris but your presentation looks strange.

isn't

x = 0 + (pi)n  and   x = pi/2 + (pi/2)n  

really just

$$x=\frac{n\pi}{2}\qquad where \qquad n\in Z$$           

 

???      

Melody  Nov 8, 2014
 #4
avatar+87333 
+5

Yeah..... your presenttion is better....we have to be careful in these trig equations about "dividing away" solutions....it's analagous to "throwing away" roots in a polynomial.....

 

CPhill  Nov 8, 2014
 #5
avatar+92806 
0

YES WE DO  !!!    ME MORE THAN YOU !!!      LOL       :)))))

Melody  Nov 8, 2014

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